\(\int \frac {(b x+c x^2)^{3/2}}{x^{7/2}} \, dx\) [89]

Optimal result

Integrand size = 19, antiderivative size = 75 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}-3 \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \]

[Out]

-(c*x^2+b*x)^(3/2)/x^(5/2)-3*c*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))*b^(1/2)+3*c*(c*x^2+b*x)^(1/2)/x^(1/2
)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {676, 678, 674, 213} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=-3 \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}} \]

[In]

Int[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

(3*c*Sqrt[b*x + c*x^2])/Sqrt[x] - (b*x + c*x^2)^(3/2)/x^(5/2) - 3*Sqrt[b]*c*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]
*Sqrt[x])]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 678

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac {1}{2} (3 c) \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx \\ & = \frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+\frac {1}{2} (3 b c) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx \\ & = \frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}+(3 b c) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right ) \\ & = \frac {3 c \sqrt {b x+c x^2}}{\sqrt {x}}-\frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}}-3 \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left ((b-2 c x) \sqrt {b+c x}+3 \sqrt {b} c x \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{x^{3/2} \sqrt {b+c x}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(7/2),x]

[Out]

-((Sqrt[x*(b + c*x)]*((b - 2*c*x)*Sqrt[b + c*x] + 3*Sqrt[b]*c*x*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(x^(3/2)*Sqrt
[b + c*x]))

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\left (2 c x \sqrt {b}\, \sqrt {c x +b}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c x -b^{\frac {3}{2}} \sqrt {c x +b}\right ) \sqrt {x \left (c x +b \right )}}{x^{\frac {3}{2}} \sqrt {c x +b}\, \sqrt {b}}\)	\(68\)
risch	\(-\frac {b \left (c x +b \right )}{\sqrt {x}\, \sqrt {x \left (c x +b \right )}}+\frac {c \left (4 \sqrt {c x +b}-6 \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )\right ) \sqrt {c x +b}\, \sqrt {x}}{2 \sqrt {x \left (c x +b \right )}}\)	\(71\)

[In]

int((c*x^2+b*x)^(3/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

(2*c*x*b^(1/2)*(c*x+b)^(1/2)-3*arctanh((c*x+b)^(1/2)/b^(1/2))*b*c*x-b^(3/2)*(c*x+b)^(1/2))*(x*(c*x+b))^(1/2)/x
^(3/2)/(c*x+b)^(1/2)/b^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\left [\frac {3 \, \sqrt {b} c x^{2} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} {\left (2 \, c x - b\right )} \sqrt {x}}{2 \, x^{2}}, \frac {3 \, \sqrt {-b} c x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} {\left (2 \, c x - b\right )} \sqrt {x}}{x^{2}}\right ] \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/2*(3*sqrt(b)*c*x^2*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*sqrt(c*x^2 + b*x)*(2
*c*x - b)*sqrt(x))/x^2, (3*sqrt(-b)*c*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + sqrt(c*x^2 + b*x)*(2*c*
x - b)*sqrt(x))/x^2]

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {7}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**(3/2)/x**(7/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**(7/2), x)

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {7}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(7/2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.75 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\frac {\frac {3 \, b c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {c x + b} c^{2} - \frac {\sqrt {c x + b} b c}{x}}{c} \]

[In]

integrate((c*x^2+b*x)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

(3*b*c^2*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) + 2*sqrt(c*x + b)*c^2 - sqrt(c*x + b)*b*c/x)/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{7/2}} \,d x \]

[In]

int((b*x + c*x^2)^(3/2)/x^(7/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(7/2), x)